Question

Solve ${\int }_0^2\frac{5x+2}{x^2+4}dx$

Answer 1

Consider the given integral.

$I={\int }_0^2\frac{5x+2}{x^2+4}dx$

$I=5{\int }_0^2\frac{x}{x^2+4}dx+{\int }_0^2\frac{2}{x^2+4}dx$

$I=I_1+I_2$

Now,

$I_1=5{\int }_0^2\frac{x}{x^2+4}dx$

Let $t=x^2+4$

$\frac{dt}{dx}=2x+0$

$\frac{dt}{2}=xdx$

Therefore,

$I=\frac{5}{2}{\int }_4^8\frac{1}{t}dt$

$I=\frac{5}{2}{\left[{\log }_e\left(t\right)\right]}_4^8$

$I=\frac{5}{2}({\log }_{e}8-{\log }_{e}4)$

$I=\frac{5}{2}\left({\log }_{e}2\right)$

Now,

$I_2={\int }_0^2\frac{2}{x^2+2^2}dx$

We know that

$\int \frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)$

Therefore,

$I_2=\frac{2}{2}{\left[{\tan }^{-1}\left(\frac{x}{2}\right)\right]}_0^2$

$I_2={\tan }^{-1}\left(\frac{2}{2}\right)-{\tan }^{-1}\left(0\right)$

$I_2={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(0\right)$

$I_2={\tan }^{-1}\left(\tan \frac{\pi }{4}\right)-{\tan }^{-1}\left(\tan 0\right)$

$I_2=\frac{\pi }{4}$

Therefore,

$I=I_1+I_2$

$I=\frac{5}{2}\left({\log }_{e}2\right)+\frac{\pi }{4}$

Hence, this is the answer.